Re: can anyone proove this?

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Posted by Guru on May 25, 19100 at 14:09:28:

In Reply to: can anyone proove this? posted by alison on June 02, 1999 at 23:27:24:

a+b+c=pi
say LHS = sin2a+sin2b+sin2c

consider sin2a = sin2(pi-(b+c))=sin(2pi - 2(b+c))
=-sin(2b+2c)
LHS=sin2b+sin2c-(sin2bcos2c+cos2bsin2c)
=sin2b(1-cos2c)+sin2c(1-cos2b)
=sin2b.2.sinc.sinc+sin2c.2sinb.sinb
=4sinb.cosb.sinc.sinc+4sinc.cosc.sinb.sinb
=4sinbsinc(cosb.sinc+cosc.sinb)
=4sinb.sinc.sin(b+c)
=4.sinb.sinc.sin(pi-a)
=4sina.sinb.sinc

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Comments: : a+b+c=pi : say LHS = sin2a+sin2b+sin2c : consider sin2a = sin2(pi-(b+c))=sin(2pi - 2(b+c)) : =-sin(2b+2c) : LHS=sin2b+sin2c-(sin2bcos2c+cos2bsin2c) : =sin2b(1-cos2c)+sin2c(1-cos2b) : =sin2b.2.sinc.sinc+sin2c.2sinb.sinb : =4sinb.cosb.sinc.sinc+4sinc.cosc.sinb.sinb : =4sinbsinc(cosb.sinc+cosc.sinb) : =4sinb.sinc.sin(b+c) : =4.sinb.sinc.sin(pi-a) : =4sina.sinb.sinc

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